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# CALCULUS II MA 114

UK

GPA 3.54

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This 12 page Class Notes was uploaded by Kennith Herman on Friday October 23, 2015. The Class Notes belongs to MA 114 at University of Kentucky taught by Staff in Fall. Since its upload, it has received 23 views. For similar materials see /class/228154/ma-114-university-of-kentucky in Mathematics (M) at University of Kentucky.

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Date Created: 10/23/15

MA 114 Recitation Sequences and Series Sequences and Series are often referenced as the most dif cult topic in Calculus 2 The reason why many students have trouble with this topic is the reliance on mathematical proof These problems are less about computation and more about understanding the mathematical ideas behind computations To succeed at these topics a student should spend a signi cant amount reviewing the material A successful student will understand the bigger ideas at play This material becomes much much much easier when you have a true understanding of what is going on math ematically Finally you should do as many example problems as pos sible These problems can come from WebWork the book suggested problems given by your TA Lecturer other books or other sources 1 Sequences In an informal sense we can think of a sequence an as being a list of numbers 017027037 39 quot 7an7an17 39 quot While this informal idea is good for gaining an initial understanding of sequences a more rigorous de nition is required Therefore a sequence an is a function f of real numbers whose domain are a set of integers That is an and we refer to the integer n as the index Now for a few examples 1 Write the rst four terms of the following sequences nl a an g 1 b bn Called the harmonic sequence n 27L 0 0n 471 TL 3 Solutions 2 6 24 a 1 2 7a These fractions can be reduced 1 1 1 b 1 lt gt 2 8 2 2 8 C 7 737 Determine the general form of the following sequences That is write the formula for the nth terrn 311717171739 39 39 1 1 1 17i767 739u 1 1 1 1 C 77 7 7quot39 3 9 27 81 Solutions a 1 1 1 b a c 1 3n Now we will discuss the idea of convergence of a sequence This is very similar to the idea of convergence of functions MA 113 topic However we de ne limits in a much more precise rnanner De nition 11 A sequence an converges to a limit L lim an L 7L4gtOO if for every 8 gt 0 there is some positive integer M such that an L lt E for allngt M In fact this is the de nition of limits used in more general and abstract rnathernatics Now we will work a few examples using this method 1 D Find the limit of bn Show that an converges to 0 Solution To use the 5 de nition of sequential limits we must nd an M which implies what we want to show That is given an arbitrary 5 gt 0 1 an 0an lt5 n for all n gt M Notice that if n gt M then i lt So what happens if we say 1 n2 M2 M We would have E 1 1 1 1 Ian 0lanlgltm 5 x 8 for all n gt M Therefore since we found an M which works for any 5 gt 0 we have proven that li1n 2 0 n4gtltgto 11 TL 271 4 Solution Before we begin using the 5 de nition of a limit we rst need to nd a candidate for the limit So we will rely on our previous Calculus knowledge to determine the limit and then prove our answer So we have 2 22 lirn lirn 4 i lirn n4gtoo 2n 4 n4gtltxgt 2 E n4gtoo Now we will use the 5 de nition to prove this answer We want to nd an M such that for any 5 gt 0 we have n 1 n n2 2 lt5 2n4 2 2n4 2n4 for all n gt M Note that 2 2 2 1 39f ltMl n4lt2n nSo1wee 8 weget n 1 n n2 2 1 1 1 ltltg 2n4 2 2n4 2n4 n M for all n gt M so we have proven our answer 1 1 Why did we choose M Consider that we wanted lt 5 a n 1 which is equivalent to 1 lt an which is again equivalent to lt n So then if we let M i it is necessarily true that any n gt M will imply lt 5 If the 5 proofs for limits do not come naturally to you don t worry With time and practice they become easier If you choose to continue on in math beyond the Calculus sequence you will see these again in Real Analysis also known as Advanced Calculus Exercises Here are some exercises to give you more practice with these limits Determine whether or not the sequence converges If it converges prove it with 5 de nition 1 1 n a 211 3 2 TL 2 n a n2 2 TL 3 n 1 a n1 Now we will discuss some more ideas for convergence of sequences The Limit Laws for sequences are the same as those for functions learned in Calculus 1 You should review these They are listed again in Chapter 10 Section 1 Theorem 12 The Squeeze Theorem Giuen sequences an bn en such that there exists an M E N This notation means that M is a nat ural number such that an 3 bn S cn for all n gt M and lim an 7L4gtOO lim cn L then lim bn L n4gtoo n4gt00 The Squeeze Theorem can be very useful for calculating limits Con sider the following example 1 Example Use the squeeze theorem to nd the limit of an n Solution 1 1 1 Consider the sequences and Notice that for all n 2 2 S 1 1 n n2 1 n2 S We showed earlier in this document that lim 0 and you 2n n Hoo n2 1 showed or should have shown in Worksheet 20 that lim 0 n4gtltxgt n 1 Therefore by the Squeeze Theorem lim 0 n4gtltxgt n Exercises Show the following limits using the squeeze Theorem 2 1 n2 1 n 2 Q n Another important theorem for sequences Theorem 13 The Monotone Bounded Convergence Theorem If a sequence is always increasing and bounded above then it conuerges If a sequence is always decreasing and bounded below then it conuerges Here are some examples and exercises 4 1 Show that an n converges using the Monotone Bounded n Convergence Theorem Solution First we show that it is monotone Speci cally we show that the sequence is always increasing We can do this two ways Take the derivative or compare adjacent terms of the sequence The derivative method Note that an has a corresponding contin uous function f f fg Taking the derivative yields f gt 0 for x gt 0 So f is increasing which means an is increas ing Adjacent terms method We want to show that anH 2 an We will do this by showing that anH an gt 0 We have 4n4 4n 12n32 an an T1 n9 n8 n8n9 Now we will show that an is bounded Note that lt 1 There 4n n 8 Thus the sequence converges fore we have the bound lt 4 2 Exercise Show that converges 4712 9112 13 converges 7 2 3 Exercise Show that ijw 2 Series A series is an in nite sum of the type SZan lim Sn N H00 where SnZana1a2an n0 are the partial sums The sequences an and Sn help to de ne the series When the limit of the partial sums Sn exists and is nite the series is said to con verge The series then is equal to this single nite value Series are not themselves sequences though they are described in terms of sequences We now list a series of Theorems and De nitions which are important for understanding series Theorem 21 Linearity of In nite Series If Zan and an both converge then 2an l bn and Z 0 an also converge with ZanianZanbn cZan Zcan c a constant De nition 22 A geometric series is a series de ned by the geometric sequence an cr where r 0 and c 0 Provided the series begins at n 0 it is as follows Zcr ccrcr2cr3cr4 n0 Theorem 23 The geometric series 2 cr has the following sum nM f or iflr lt 1 nM loo ifr 2 1 Theorem 24 Divergence Test If a series Zan converges then limrH00 an 0 In other words if the limH00 an y 0 then 2 an diverges Examples and Exercises 1 Telescoping series Determine the sum of the following series using 2 the limit of the partial sums E m n nl Solution First notice that the series does not fail the divergence test 352 W2 1 i 0 So we will look at the partial sums of the series However rst note that 4n2 l 2n l2n 1 Using partial fractions decomposition left for the reader to show we have that 2 i 1 1 ITZn 1 2nf 7 D 00 Determine whether 2 7 n n70 Now to gain some intuition consider some of the rst few partial sums 32 1 l 2 1 33 ll W W i 1 S4 1 33 77 1 Note that all of the middle terms cancel one with one another This gives us the general formula for Sn l So now we 11 take the limit of the partial sums lim 1 1 new 271 1 Thus 7 3 converges or diverges If it con verges nd the sum 7 Solution First apply the divergence test 3 1 7 0 139 1m 7 new 7713 4 7 The series diverges 0 1 TL Determine whether 2 converges or diverges If it con n70 verges nd the sum Solution Notice that this series is a geometric series with r Thus 3 n0 3 H 0 m 51 00 1 Exercise Determine whether conver es or R2 1 g diverges If it converges nd the sum xerc1se e erm1ne w e er conver es or 1ver es 1 E 39 D t 39 h th 2 g d39 g If 39t n0 converges nd the sum 00 9n xerc1se e erm1ne w e er converges or 1verges E D t h th 2 7 2 d n n0 If it converges nd the sum Exercise Determine whether i 1 converges or di 4 2 39 TF2 n 3n 1 verges If it converges nd the sum 00 e n xerc1se e erm1new e er conver es or 1ver es E Dt hth g d g If 7T nl it converges nd the sum 00 7T n ExerCISe Determine whether 2 converges or diverges If n4 it converges nd the sum Series Convergence For this section we will list each convergence test in the form of a Theorem Then follow it up with an two examples At the end of this section we will provide a plethora of exercises At this points in the notes there is a slight divergence from the format of the book I will list all of the information for series conver gence in slightly different but logical order 31 Convergence of Positive Series Theorem 31 The Integral Test Let an where f is a contin uous decreasing and positive function Then i f fxdac converges then Zan converges 1 nl ii If fxdac diverges then Zan diverges 1 n1 1 Example 32 Determine whether the series 2 converges or di n nl verges Solution We will apply the Integral Test Consider the function f We need to ensure that this function is continuous positive and decreasing We know that the function is all ts all of these criteria However you should practice showing that the function is decreasing Now consider the integral 1 R 1 R dx lim dx lim lnac lim lnR I gt oo 1 f R4gtltxgt 1 f R4gtltxgt 1 R4gtltxgt 1 Therefore 2 diverges by the Integral Test D n nl 2 Example 33 Determine whether the series 2 2 converges or di n nl verges using the Integral Test 2 Solution We will apply the Integral Test The function f 2 is as continuous positive and decreasing Why is this true Show these steps yourself You will be responsible for arguing why these things are true in an exam scenario Now R 2 R 2 2 2 2dsc lim 2dsc lim lim 22 1 R4gtltXgt 1 f R4gtltxgt a 1 T4gtOO R 10 2 Therefore the series 2 2 converges by the Integral Test D n nl The following theorem is a essentially a corollary of the lntegral Test 1 Theorem 34 The serles 2 E converges lf and only lfp gt 1 nl Using this information Example 33 would simply be complete by referencing this Theorem Theorem 35 Comparison Test leen sequences ambn such that an 3 bn for all n 239 if bn converges then Eon converges ll IfZan dlverges then an dlverges 7 Example 36 Determine whether the series 2 converges or di n nl verges 1 7 1 Solutlon Notice that lt and we showed previously that Z n n n nl diverges Therefore this series diverges by the comparison test B 0 Example 37 Determine whether the series 2 nl converges or 16n2 diverges 1 2 2 Solutlon Notice that W lt E and we previously showed that E E is convergent Thus this series is convergent D Theorem 38 Limit Comparison Test If an and bn are posltlve sequences and the llmlt lim b n L exlsts then n4gt00 o fL gt 0 then Zan converges lf and only of bn converges ll

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